Consider the following reactions: (I) CH3 - CH = CH2 (i) B2H6/THF → (ii) H2O2/OH^- (A)

Question

Consider the following reactions:

(I) \(CH_3 - CH= CH_2 \xrightarrow [(ii)\,H_2O_2/OH^-]{(i)\,B_2H_6 /THF}(A)\)

(II) \(CH_3-CH= CH_2 \xrightarrow{H^+/H_2O} (B)\)

Identify the major product(s) (A) and (B) formed in the above reactions.

(1) \((A) : CH_3 -\overset {OH}{\overset{|}{C}H} - CH_3\)  ;  \((B) : CH_3 -\overset {OH}{\overset{|}{C}H} - CH_3\)

(2) \((A): CH_3 - CH_2 - CH_2 - OH\)  ;  \((B) : CH_3 -\underset{OH}{\underset{|}{C}H} -CH_3\)

(3) \((A) : CH_3 -\overset {OH}{\overset{|}{C}H} - CH_3\)  ;  \((B): CH_3 - CH_2 - CH_2 - OH \)

(4) \((A): CH_3 - CH_2 - CH_2 - OH \)  ;  \((B): CH_3 - CH_2 - CH_2 - OH \)

Answer 1

(2) \((A): CH_3 - CH_2 - CH_2 - OH\) ; \((B) : CH_3 -\underset{OH}{\underset{|}{C}H} -CH_3\)

Hydroboration of unsymmetrical alkene followed by hydrolysis by alkaline H₂O₂ results in the formation of a product as if H₂O has been added to alkene according to anti-Markovnikov rule.

Acid catalysed addition of water to unsymmetrical alkene follows Markovnikov rule with the possibility of rearrangement.